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Representing shm with circular motion - class-XI

Description: Representing shm with circular motion
Number of Questions: 15
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Tags: simple harmonic motion physics oscillations and waves oscillations
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A force acts on a $30$gm particle in such a way that the position of the particle as a function of time is given by $x=3t-4t^2+t^3$, where x is in metres and t is in seconds. The work done during the first $4$ second is?

  1. $2.88$J

  2. $450$mJ

  3. $490$mJ

  4. $530$mJ


Correct Option: A
Explanation:

$M=20gm=0.020kg$

$x=3t-4t²+t³m$
$v=dx/dt=3-8t+3t²m/s$
$a=dv/dt=-8+6tm/s²$
$Force=ma=0.120t-0.160N$
$Workdone=dW=Fdx=F\ast \frac { dx }{ dt } \ast dt=(0.120t-0.160)\ast (3-8t+3t²)dt=0.360t³-1.440t²+1.640t-0.480$
$W=0.090t⁴-0.490t³+0.82t²-0.480t=2.88J$

A wheel of radius $1$ meters rolls forward half a revolution on a horizontal ground. The magnitude of the displacement of the point of the wheel initially in contact with the ground is:

  1. $2 \pi$

  2. $\sqrt 2 \pi$

  3. $\sqrt {{\pi ^2} + 4} $

  4. $\pi$


Correct Option: C
Explanation:
Let fro $A$ to $B$, where complete half a rotation.
we have,
$PQ=\alpha =$ half circumference $=\pi \ m$
and $P' Q=$ diameter $=2\ m$
so we have displacement of $P$
$x=|PP;|=\sqrt {PQ^2 +P; Q^2}$ (Pythagoras theorem)
$\Rightarrow \ x=\sqrt {\pi^2 +4}m$

Two SHMs are represented by the equations 
$y1=10sin(3\Omega t+\frac{\Omega }{4})$ and
$y2=5[sin3\Omega t+\sqrt{3}cos 3\Omega t]$. their amplitudes and in the ratio

  1. 1:2

  2. 2:1

  3. 1:3

  4. 1:1


Correct Option: B
Explanation:

Compare the given equations with,

$y=a\sin \omega t$

Where a is amplitude of wave

Here, a1 = 10

 a2 = 5

So the ratio is

$\dfrac{{{a} _{1}}}{{{a} _{2}}}=\dfrac{10}{5}=\dfrac{2}{1}$

Or $2:1$

A simple harmonic oscillator starts from extreme position and covers a displacement half of its amplitude in a time '$t$', the further time taken by it to reach mean position is

  1. $2t$

  2. $t$

  3. $ t/\sqrt{2} $

  4. $t/2$


Correct Option: D
Explanation:

$x=A cos(\omega t)$
Taking $t=0$ at extreme position.
$x=\dfrac{A}{2}$ is reached in time t
$\omega t=\dfrac{\pi }{3} ;\omega =\dfrac{\pi }{3t}$
$x=0$ is reached at $\omega t _{1}=\dfrac{\pi }{2}$
$t _{1}=\dfrac{\pi }{2\omega }=\dfrac{\pi \times 3t}{2\times\pi }=\dfrac{3t}{2}$
$t _{1}-t=\dfrac{t}{2}$

The circular motion of a particle whose speed is constant is

  1. Periodic but not simple harmonic

  2. Simple harmonic but not periodic

  3. Periodic and simple harmonic

  4. Neither periodic not simple harmonic


Correct Option: A
Explanation:

Uniform circular motion is a periodic motion but not simple harmonic

The correct option is (a)

Find the distance covered by a particle from time $t=0$ to $t=6\ \sec$. When the particle followsa the movement in straight line according to $y=a\cos \left(\dfrac {\pi}{4}\right)t$:-

  1. $a$

  2. $2a$

  3. $3a$

  4. $4a$


Correct Option: C

The particle executes SHM on a straight line. At two positions its velocity $u$ and $v$ while acceleration, $\alpha$ and $\beta$ respectively $[\beta > \alpha >0]$, the distance between the two positions will be:-

  1. $\dfrac{u^2+v^2}{\alpha+\beta}$

  2. $-\dfrac{u^2-v^2}{\alpha+\beta}$

  3. $\dfrac{u^2-v^2}{\alpha-\beta}$

  4. $\dfrac{u^2+v^2}{\beta-\alpha}$


Correct Option: B
Explanation:

We know that velocity is related as,

$u=\omega \sqrt{A^2-X _1^2}$
$v=\omega\sqrt{A^2-X _2^2}$
Acceleration is related as,
$\alpha=-\omega^2X _1^2$
$\beta=-\omega^2X _2^2$
$\dfrac{u^2-v^2}{u^2}=-(X _1^2-X _2^2)$
$=-(X _1+X _2)(X _1-X _2)$
$=\dfrac{\alpha+\beta}{u^2}(X _1-X _2)$
$X _1-X _2=-\dfrac{u^2-v^2}{\alpha+\beta}$

A ball is projected from the ground at angle 0 with the horizontal. After 1 sec it is moving at angle ${ 45 }^{ \circ  }$ with the horizontal and after 2s it is moving horizontally. What is the velocity of projection of the ball ? (Take $g=10\quad { ms }^{ -2 }$)

  1. $10\sqrt { { 3ms }^{ -1 } } $

  2. $20\sqrt { { 3ms }^{ -1 } } $

  3. $10\sqrt { { 5ms }^{ -1 } } $

  4. $20\sqrt { { 2ms }^{ -1 } } $


Correct Option: C

To understand Simple Harmonic Motion as analogous to circular motion,

  1. we project the circular motion of the particle along any radius.

  2. we project the circulation motion of the particle along a chord.

  3. we project the circulation motion of the particle along the diameter.

  4. None of these.


Correct Option: C
Explanation:

The circular motion can be resolved into a S.H.M. In this process we project circular motion of particle along diameter the projection of particle performs S.H.M. with centres as mean positions.

The actual distance moved along the circle will be                   the distance moved by the projection on the diameter.

  1. less than

  2. equal to

  3. greater than

  4. None of these


Correct Option: C
Explanation:

Let the radius of the circle be $R$.

Thus distance moved along the circle  $D = \pi R = 3.14 R$
Projection of this distance along the diameter is equal to the diameter of the circle i.e.  $2R$
Thus actual distance moved along the circle will be greater than the distance moved by the projection on the diameter.

The particle executes SHM on a straight line. At two positions its velocity $u$ and $v$ while acceleration, $\alpha$ and $\beta$ respectively $[\beta > \alpha >0]$, the distance between the two positions will be:-

  1. $\frac{u^2+v^2}{\alpha+\beta}$

  2. $-\frac{u^2-v^2}{\alpha+\beta}$

  3. $\frac{u^2-v^2}{\alpha-\beta}$

  4. $\frac{u^2+v^2}{\beta-\alpha}$


Correct Option: B
Explanation:
$u=w\sqrt{A^2-X _1^2}$
$V=w\sqrt{A^2-X _2^2}$
$\alpha=-w^2 x _1^2$
$\beta=-w^2 x _2^2$
$\frac{u^2-v^2}{u^2}=-(x _1^2-x _2^2)$
$=-(x _1+x _2)(x _1-x _2)$
$=\frac{(\alpha+\beta)}{w^2}(x _1-x _2)$
$x _1-x _2=\frac{u^2-v^2}{\alpha+\beta}$

A stone is projected from the ground with a velocity of $14 \ ms^{-1}$ one second later it clean a wall $2 \ m$ high. The angle of projection is $(g = 10 \ ms^{-2})$ 

  1. $30 ^\circ$

  2. $45 ^\circ$

  3. $60 ^\circ$

  4. $15 ^\circ$


Correct Option: A

The period of a particle it is $8s$. At $t=0$ it is at the mean position. The ratio of the distance covered by the particle in first second and second will be

  1. $\cfrac { \sqrt { 2 } -1 }{ \sqrt { 2 } } $

  2. $\cfrac { 1 }{ \sqrt { 2 } } $

  3. $\cfrac { 1 }{ \sqrt { 2 } -1 } $

  4. $\left[ \sqrt { 2 } -1 \right] $


Correct Option: C
Explanation:

$x=A\sin\omega t$

$T= 8 sec$
$\dfrac{2\pi}{\omega}=8 sec$
$\omega=\dfrac{2\pi}{8}=\dfrac{\pi}{4}$
$x=A\sin \dfrac{\pi}{4} t$ ,At $t=1 sec$
$x _1=\dfrac{A}{\sqrt2}$ ,At $t=2sec$
$x _2=A$
Required ration $=\dfrac{x _1}{x _2-x _1}=\dfrac{1/\sqrt2}{1-\dfrac{1}{\sqrt2}}=\dfrac{1}{\sqrt2-1}$

Sunlight of intensity 1.3 kW m–2 is incident normally on a thin convex lens of focal length 20 cm. Ignore the energy loss of light due to the lens and assume that the lens aperture size is much smaller than its focal length. The average intensity of light, kW m–2, at a distance 22 cm from the lens on the other side is _________. 

  1. $ 130 $

  2. $120 \ $

  3. $ 170$

  4. $160 $


Correct Option: A,C

Simple harmonic motion is the projection of uniform circular motion on the

  1. $x$- axis

  2. $y$- axis

  3. reference circle

  4. any diameter of reference circle.


Correct Option: D
Explanation:

Simple harmonic motion is the projection of uniform circular motion on any diameter of reference circle.

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